E 600 mm B Rigid P. 80 mm 160 mm 80 mm

Q: Determine the normal stress at corners A and B of the column.

A: Draw the figure.

Q: 3 ft- Wo G 4 ft 60% OD MG 2 ft FG (NG B G E B.

A:

Q: N 20 mm 200 mm 20 mm 300 mm pmri 20 mm A 1.5 m C N N 站 B D TTTTTTT 1.5 m 5m

A: given; load at point c (Pc)=430Nload at point D(PD)=634NAC=1.5mBD=1.5mAB=5mCD=2mlets take reaction…

Q: 25 lb 40 lb 24 in. 0.5 in. 4 in. a B Section a – a

A:

Q: 200 mm C 600 mm 200 mm

A: Area moment of inertia is a geometrical property that defines how different points are arranged with…

Q: 330 mm 220 100 mm mm 800 N 75 mm 200 N 10 1200 N

A:

Q: 75 mm D. 1.5 m E F B 2 m 150 mm 5 deg 2 deg 4 deg 1.5 deg 1 deg 3.5 deg

A: Write the twist angle of point A respect to point B.   φA/B=φA-φBφA-φB=3°…

Q: 250 mm K3 rad/S 300 300 mm 400 mm

A: AS PER THE POLICY OF THE COMPANY, ONLY FIRST 3 SUBPARTS WILL BE ANSWERED. KINDLY POST REST OF THE…

Q: Fx A 75 mm 50 mm 150 mm 350 mm В F2 250 mm y

A:

Q: Vertically 300lb/pulg2 and horizontally 150 lb/pulg2 I have only this data please answer.

A: Giver:- My=300 lb/pulg2 Mx=150 lb/pulg2   To find:- Resultant

Q: F 1000 lb y 500 lb 45° 30°

A:

Q: Cl 3. F (9001) O 0 P d DO

A:

Q: -150 mm 15 mm B C 150 mm -15 mm 15 mm - 100 mm·

A: given;

Q: w = 35 kN/m 38-mm diameter 38-mm diameter 150-mm diameter 100 mm 300 mm 100 mm

A:

Q: 25 mm R50mm 100 mm

A: Given

Q: erage 1.5 kN 1.5 kN A B Uniform Thickness-Smm 20 mm 0.4 m 0.4 m 0.2 m 20 mm

A:

Q: в 140 MPa 80 MPa a) 0,011 mm b) -0,011 mm c) 0,021 mm d) -0,021 mm e)-0,019 mm

A:

Q: - 100 mm 20 mm yo 140 mm -20 mm 20 mm - 100 mm

A: Rectangle Area (mm2) x¯ (in) y¯ (in) x¯y¯A (in4) I 2000 -50 70 -7000000 II 2400 0 0 0 III 2000…

Q: 30 mm OF 10 mm 30 mm 10 mm X

A: Option C is correct 503675.34 mm4

Q: 50 mm 150 mm 150 mm 300 mm *

A: A1=300×50A1=15000 mm2y¯1=3002=150 mm   from bottom,A2=300×50A2=15000 mm2y¯2=300+502=325 mm,

Q: 22. Temporary joint is obtained by one of the following. (A) Interference fit (B) welding (C)…

A: Temperory joint is obtained by threaded connections.

Q: Y. 20 mm 20 mm 260 mm 20 mm 200 mm

A:

Q: 600 mm C 360mm

A:

Q: a) 10mm IG mm IG mm 1050mm IG mm 16mm 250 mm 300 mm

A:

Q: 15 mm 20 mm 100 mm 50 mm P. Fig. P2.97 and P2.98

A:

Q: cement: FLDE mm CF mm mm 1 1.5 84 = 8p + 8c = mm tan a = α Ans 1500 FALAH = A/H mm AH 8 = 8g +84H mm…

A:

Q: Locate the centroid of the plane area shown.

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: - B А The value of toughness is measured the under point Point D to point A B. Zero to point D Point…

A: Dear student, as per the given options there is no correct answer. Please find the explanation…

Q: 2. Hooke's law holds good upto A.Yield point B.Limit of proportionality C.Breaking point D.Elastic…

A: To explain the Hooke's law limits

Q: -r = 100 mm 50 mm -180 mm

A: As per our guidelines, experts can provide solution to only one problem in case of multiple…

Q: Bronze A = 300mm² E = 83 GPa 2.00 m Steel A = GO0mm? E = 200 GPa 1.00 m A B

A: GivenMN=45kN

Q: 120 mm 240 mm 120 mm

A: .

Q: В 125 mm A 175 mm C OD - 225 mm 150 N 75 mm

A:

Q: E= 103 GPa E = 75 GPa E= 190 GPa 0.8 mm 150 mm 100 mm 300 mm Bronze Al Steel

A: Stress produces in the material due to changes in temperature is known as thermal stress. Condition…

Q: Q16: Ferrous materials such: a: Mg ---- b: cast iron c: steel d: b+c

A: As per our guidelines, we are supposed to answer only first one question in case of multiple…

Q: - R60mm Fig 10 100mm 20hm 20mm 100mm 40mm pun ser

A: For a composite area consisting of many areas Ai, each with centroid at locations xi¯,yi¯, the…

Q: 200 mm 300 mm Steel 75 mm Brass

A: Givenρb=8.75 Mg/m3=8750 kg/m3ρs=7.87 Mg/m3=7870 kg/m3

Q: (1) (2) (3) 50 N 10 mm 10 mm 10 mm A:=50 mm? A2=20 mm? A:=10 mm? E=200 GPa

A:

Q: Determine the force in members BD and CD of the truss shown.

A: For the given truss   To determine  forces in member BD and CD

Q: Which of the following statement(s) is/are correct? i. The area under stress-strain curve gives…

A: The total area under stress strain curve is called toughness. As in (i) it is not mentioned that up…

Q: Determine the x and y components of each of the forces shown.

A: To find out the x component of the forces Resolve all the force in the x-axis…

Q: 16 KN/m 10 KNm 4m 1.35m 3.15m 1.5m 210 mm 180 mm 12 mm 9 mm 180 mm 210 mm W SEZ 9 mm 15 mm 90 mm

A:

Q: 4.5KM/M 225 mm 8m 125mm 35mm

A: Given Data: The magnitude of load, w=4.5 kN/m

Q: Match thể Items Column I P Addendum

A: Given: Match the following,

Q: M N.M A150 N 300 mm 9o N 400 mm

A: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and…

Q: 24 mm 5 mm 30 mm 8 mm 5 mm 30 mm Section AA 12,000 N

A: The correction factor is rc=3015=2 The correction factor for 2 is k=1.41. The moment of inertia is…

Q: F126KN 180KN 26 -3m 1.5m 48 65° 145KN V165 KN

A: Drawing the system of force as shown below – Let us take point A as an origin of the coordinate…

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.